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Function problem

  1. Jul 27, 2005 #1
    there's an example question in a book i'm studying
    "please solve
    (2*x to the 3rd power +y to the 3rd power)dx-3x*y to the 2nd power*dy=0"
    i have the steps to the solution but i don't know why the problem is solved in thosed steps. Can someone explain?
    p.s.
    can someone recommend a good book for math for someone in a physics major?
     
  2. jcsd
  3. Jul 27, 2005 #2
    What exactly are you meant to be solving for? Show the steps you have..
     
  4. Jul 27, 2005 #3

    HallsofIvy

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    (2x3+ y3)dx- 3xy2dy= 0

    This is a first order differential equation. It's almost exact since (2x3+ y3)y= 3y2 while (-3xy2)x= -3y2, differing only in the sign. Such an equation always has an "integrating factor"- a function such that, if you multiply the entire equation by the function the equation becomes exact but there is no general way of finding that integrating factor.

    Here, because the terms involve only powers of x and y, I decided to try an integrating factor of the form xnym. Then the test for exactness becomes (2x3+nym+ xny3+m)y= 2mx3+nym-1+(3+m)xny2+m= (-3x1+ny2+m)x= -3(1+n)xny2+m.

    That is, we must have 2mx3+nym-1+(3+m)xny2+m= -3(1+n)xny2+m.
    The second term on the left has the same powers as the right but we need to get rid of the first term- OK, take m= 0! That "kills" the first term and to make the terms left equal we need to have 3+m= -3(1+n) or 3= -3(1+n) since m=0. That gives
    1+n= -1 or n= -2.

    Let's try that: multiplying (2x3+ y3)dx- 3xy2dy= 0
    by x-2 gives (2x+ x-2y3)dx- 3x-1y2dy= 0.
    (2x+ x-2y3)y= 3x-2y2 and
    (-3x-1y2)x= 3x-2y2.
    Yes!! Those are the same so this equation is exact.

    That MEANS that there exist some function, F(x,y), such that dF= (2x+ x-2y3)dx- 3x-1y2dy or, same thing, that
    Fx= 2x+ x-2y3and Fy= - 3x-1y2 (you might remember that from multi-variable calculus).

    Since Fx= 2x+ x-2y3, taking the anti-derivative (with respect to x), F(x,y)= x2- x-1y3+ g(y).
    (The partial derivative with respect to x treats y like a constant. Taking the "anti-derivative" we treat y like a constant so that "constant of integration" may be any function of y.)

    From that, Fy= -3x-1y2+ g'(y) and that must be equal to - 3x-1y2 . Comparing the two, that tells us that g'(y)= 0 or that g really is a constant.

    IF F(x,y)= x2- x-1y3+ C, for any constant C, then
    dF=(2x+ x-2y3)dx- 3x-1y2dy = 0. Since dF= 0, F is actually a constant: F(x,y)= x2- x-1y3+ C= C' where C' is also a constant, not necessarily equal to C. Since subtracting C from C' will still be an unknown constant, we can just write that as "C" and have
    x2- x-1y3= C as the general solution to the differential equation.
     
  5. Jul 27, 2005 #4
    Another line of reasoning is:
    (2x3+ y3)dx- 3xy2dy= 0

    looking at the differential equation we see that it is
    homogeneous (F(tx, ty) = tnF(x,y)). In this case,
    each of the terms has degree 3.
    So let's do a change of variables y = vx.

    (2x3 + (vx)3)dx - 3x(vx)2(vdx + xdv) = 0
    which gives us

    x3( (2+v3) - 3v2(v + xv')) = 0
    2 + v3 - 3v3 - 3xv2v' = 0
    2(1 - v3) - x 3v2 v' = 0

    which is seperable:
    2 (dx/x) = 3v2 dv/(1 - v3)

    2 d(ln(x)) + d( ln(1-v3)) = 0
    2 ln(x) + ln(1-v3) = C'

    ln( x2(1-v3) ) = C'
    x2(1-v3) = C

    switching back to y:
    x2(1- (y/x)3) = C
    x2- x-1y3 = C

    as before.
     
    Last edited: Jul 27, 2005
  6. Jul 28, 2005 #5
    How do you know to make the change of variables y = vx?
     
  7. Jul 28, 2005 #6

    HallsofIvy

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    That's a standard substitution for homogeneous equations. The point is that in a homogeneous equation we can always rewrite the equation so x and y only appear as [tex]\frac{y}{x}[/tex].

    I missed that- qbert's solution is much simpler than mine!
     
    Last edited: Jul 28, 2005
  8. Jul 28, 2005 #7
    but why would someone think of to make that substitution?
     
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